Continuing my search of an integral expression for the sequence:
$$a_s=\sum_{n=0}^s {s \choose n} \frac{B_n}{n!}$$
(Where $B_n$ is the $n$th Bernoulli Number, $B_1$ is taken to be $\frac{1}{2}$ and $s$ must be a positive integer, see my other post) I came across the following development, which may be incorrect, since there is numerical evidence to think so. I would be interested in finding the flaw and trying to correct it (if possible) to get a correct result.
Let's start by pointing out the fact that, by the definition of binomial coefficient:
$$\sum_{n=0}^s {s \choose n} \frac{B_n}{n!}=\sum_{n=0}^\infty {s \choose n} \frac{B_n}{n!}$$
Then, since:
$${s \choose n}= \frac{1}{2\pi} \int_{-π}^π e^{itn}(1+e^{it})^s \, dt$$
We have that:
$$\frac{1}{2\pi} \sum_{n=0}^\infty \int_{-π}^π e^{itn}(1+e^{it})^s \, dt \frac{B_n}{n!}$$
Up to here, everything seems to work fine.
Then, as the sum is absolutely convergent for positive integer $s$, say
$$\frac{1}{2\pi} \sum_{n=0}^\infty \left | \int_{-π}^π \frac{B_n}{n!} e^{itn}(1+e^{it})^s \, dt \right | <\infty$$
We are allowed to interchange the sum and the integral operators (by the Fubini/Tonelli theorems, see this other question) to get:
$$\frac{1}{2\pi} \int_{-π}^π (1+e^{it})^s \sum_{n=0}^\infty e^{itn} \frac{B_n}{n!} \, dt$$
By the Generating Function of Bernoulli Numbers, and since for any complex $z$ and integer $n$, $e^{zn}=(e^z)^n$,
$$\sum_{n=0}^\infty e^{itn} \frac{B_n}{n!}=\frac{e^{it}}{1-e^{-e^{it}}}$$
So that our original function/sequence can be expressed as:
$$\frac{1}{2\pi} \int_{-π}^π (1+e^{it})^s \frac{e^{it}}{1-e^{-e^{it}}} \, dt$$
The problem is that, fom example, setting $s=3$ in our original sequence (the sum), we would get $a_3=\frac{11}{4}$, but the integral will equal $1$. This error continues for all other values of $s$ computed.
Any help with this? Where is the flaw and how could it be solved?
Thank you very much.
As stated in the comments, the correct integral representation for the binomial coefficients is given by: $$ {s \choose n}= \frac{1}{2\pi} \int_{-π}^π e^{-itn}(1+e^{it})^s \, dt $$ With this change, all of the results are accurate.